I've been working with symbolic operations in a robotics class and we've been having trouble getting concise results. First, the results were given in gigantic fractions, but managed to resolve this using a combination of both simplify() first, and then vpa(,2).
What is now a problem is that the results are really long. They have a bunch of terms (sine, cosine, all in a sum), but most of those terms are essentially zero.
Here is an example result in a 4 x 4 matrix:
6.1? โ 17 โ ?3 โ ???(?โ1) + ?2 โ ???(?โ1) โ ???(?โ2) + ?3 โ ???(?โ1) โ ???(?โ2) โ 6.1? โ 17 โ ?2 โ ???(?โ1) โ ???(?โ2) + 6.1? โ 17 โ ?6 โ ???(?โ1) โ ???(?โ5) + 6.1? โ 17 โ ?3 โ ???(?โ1) โ ???(?โ2) + 3.7? โ 33 โ ?3 โ ???(?โ2) โ ???(?โ1) โ 6.1? โ 17 โ ?3 โ ???(?โ1) โ ???(?โ2) + 2.3? โ 49 โ ?6 โ ???(?โ4) โ ???(?โ5) โ ???(?โ1) โ 1.0 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ5) โ 6.1? โ 17 โ ?6 โ ???(?โ2) โ ???(?โ1) โ ???(?โ5) โ 6.1? โ 17 โ ?6 โ ???(?โ4) โ ???(?โ1) โ ???(?โ5) โ 1.0 โ ?6 โ ???(?โ5) โ ???(?โ1) โ ???(?โ4) โ 1.4? โ 65 โ ?6 โ ???(?โ1) โ ???(?โ4) โ ???(?โ5) + 1.0 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5) โ 1.4? โ 65 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5) โ 2.3? โ 49 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ5) โ ???(?โ4) + 1.4? โ 65 โ ?6 โ ???(?โ1) โ ???(?โ4) โ ???(?โ5) โ ???(?โ2) + 8.6? โ 82 โ ?6 โ ???(?โ2) โ ???(?โ4) โ ???(?โ5) โ ???(?โ1) โ 6.1? โ 17 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5) โ 3.7? โ 33 โ ?6 โ ???(?โ1) โ ???(?โ4) โ ???(?โ2) โ ???(?โ5) โ 6.1? โ 17 โ ?6 โ ???(?โ1) โ ???(?โ5) โ ???(?โ2) โ ???(?โ4) โ 2.3? โ 49 โ ?6 โ ???(?โ2) โ ???(?โ4) โ ???(?โ1) โ ???(?โ5) โ 3.7? โ 33 โ ?6 โ ???(?โ2) โ ???(?โ5) โ ???(?โ1) โ ???(?โ4) โ 6.1? โ 17 โ ?6 โ ???(?โ4) โ ???(?โ5) โ ???(?โ1) โ ???(?โ2) โ 8.6? โ 82 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5) โ 5.3? โ 98 โ ?6 โ ???(?โ2) โ ???(?โ1) โ ???(?โ4) โ ???(?โ5) + 8.6? โ 82 โ ?6 โ ???(?โ4) โ ???(?โ1) โ ???(?โ2) โ ???(?โ5) + 1.4? โ 65 โ ?6 โ ???(?โ5) โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) + 3.7? โ 33 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5)
As you can see, the result has mostly terms in a 
scale. I had seen before in different excercises that when working with symbolics, MATLAB would express a 1 (which had to be 1) as 1.00000000000, which makes me wonder if MATLAB is having trouble generating concise values when working with trigonometry.
I tried also having "pi" as symbolic, but it didn't do much to the problem at hand. Format short does nothing too.
What I would like to do:
- Ideally, I'd like for MATLAB to calculate correctly these functions. As I know it may even be impossible for the program:
- Is there a way to tell MATLAB that any terms below a certain weight/value/relevance/idk be eliminated or considered 0?
This is my most recent code with said problem, the resulting matrix Mov has the problem. The values of interest are the ones in the last row, the first 3 are not of interest. The result shown above is the first term of this matrix (1,4), which, according to me, should be:
?2 โ ???(?โ1) โ ???(?โ2) + ?3 โ ???(?โ1) โ ???(?โ2)โ 1.0 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ5) + 1.0 โ ?6 โ ???(?โ1) โ ???(?โ2) โ ???(?โ4) โ ???(?โ5)
Code:
clear all, clc, close all
syms th1 th2 th3 th4 th5 th6 L1 L2 L6 T d3 pi
Rz1=[cos(th1) -sin(th1) 0 0; sin(th1) cos(th1) 0 0; 0 0 1 0; 0 0 0 1]
Tr1=[1 0 0 0;0 1 0 0;0 0 1 L1;0 0 0 1]
Rx1=[1 0 0 0;0 cos(-pi/2) -sin(-pi/2) 0;0 sin(-pi/2) cos(-pi/2) 0;0 0 0 1]
Rz2=[cos(th2) -sin(th2) 0 0; sin(th2) cos(th2) 0 0; 0 0 1 0; 0 0 0 1]
Rx2=[1 0 0 0;0 cos(pi/2) -sin(pi/2) 0;0 sin(pi/2) cos(pi/2) 0;0 0 0 1]
Tr2=[1 0 0 L2;0 1 0 0;0 0 1 0;0 0 0 1]
Rz3=[cos(pi/2) -sin(pi/2) 0 0; sin(pi/2) cos(pi/2) 0 0; 0 0 1 0; 0 0 0 1]
Rx3=[1 0 0 0;0 cos(pi/2) -sin(pi/2) 0;0 sin(pi/2) cos(pi/2) 0;0 0 0 1]
Tr3=[1 0 0 0;0 1 0 0;0 0 1 d3;0 0 0 1]
Rx4=[1 0 0 0;0 cos(-pi/2) -sin(-pi/2) 0;0 sin(-pi/2) cos(-pi/2) 0;0 0 0 1]
Rz4=[cos(th4) -sin(th4) 0 0; sin(th4) cos(th4) 0 0; 0 0 1 0; 0 0 0 1]
Rz5=[cos(-pi/2) -sin(-pi/2) 0 0; sin(-pi/2) cos(-pi/2) 0 0; 0 0 1 0; 0 0 0 1]
Rx5=[1 0 0 0;0 cos(-pi/2) -sin(-pi/2) 0;0 sin(-pi/2) cos(-pi/2) 0;0 0 0 1]
Rz6=[cos(th5) -sin(th5) 0 0; sin(th5) cos(th5) 0 0; 0 0 1 0; 0 0 0 1]
Tr4=[1 0 0 L6;0 1 0 0;0 0 1 0;0 0 0 1]
Rz7=[cos(-pi/2) -sin(-pi/2) 0 0; sin(-pi/2) cos(-pi/2) 0 0; 0 0 1 0; 0 0 0 1]
Rx6=[1 0 0 0;0 cos(-pi/2) -sin(-pi/2) 0;0 sin(-pi/2) cos(-pi/2) 0;0 0 0 1]
Rz8=[cos(th6) -sin(th6) 0 0; sin(th6) cos(th6) 0 0; 0 0 1 0; 0 0 0 1]
Mov=vpa(simplify(Rz1*Tr1*Rx1*Rz2*Rx2*Tr2*Rz3*Rx3*Tr3*Rx4*Rz4*Rz5*Rx5*Rz6*Tr4*Rz7*Rx6*Rz8),2)
